Can an upper triangular matrix be put in jordan form using upper triangular matrices and permutations only.

Solution 1:

We bring a matrix to JNF by changing to a basis of (generalized) eigenvectors, so we want to show that such a change of basis can be implemented by an upper triangular matrix. Equivalently, we want to show that there is a basis $v_1, \ldots , v_n$ of generalized eigenvectors, such that the components of $v_j$ are zero beyond the $j$th entry.

The eigenvalues of $A$ are its diagonal entries, and if these never repeat, then this is trivial. So the only challenge are multiple eigenvalues, and let's say we have a block that looks like $$ B = \begin{pmatrix} a & * & * \\ 0 & a & * \\ 0 & 0 & a \end{pmatrix} , $$ and $$ A = \begin{pmatrix} A_0 & * & * \\ 0 & B & * \\ 0 & 0 & A_1 \end{pmatrix} , $$ with the $A_j$'s upper triangular also. Of course, in the actual matrix, these entries $a$ don't need to follow each other, but for this, we can make use of the permutation matrix.

Again, it's clear that $(B-a)^j e_j = 0$ for $j=1,2,3$, and by giving these vectors suitable earlier components, we obtain generalized eigenvectors of $A$ also.

Solution 2:

It's not true for sure for matrices of dimension $n\times n$ with $n\ge 6$. If it was the case, for any upper triangular unipotent matrix $A$ you would have that for some upper triangular $U$ the matrix $UAU^{-1}$ belongs to the finite set of matrices of the form $P^{-1}JP$, where $P$ is a permutation matrix and $J$ a Jordan matrix. But there are infinitely many orbits of conjugation action of upper triangular matrices on nilpotent upper triangular matrices. See Djoković, Malzan Orbits of Nilpotent Matrices https://www.sciencedirect.com/science/article/pii/0024379580900142

You can also take the following $4\times 4$ matrix (example based on the above paper):

$$\begin{pmatrix} 0 & 1 & 0 & 0 \\ & 0 & 0 & 1 \\ & & 0 & 1 \\ & & & 0 \end{pmatrix}$$

and prove by hand it can't be brought to Jordan form by upper triangular matrices.